Q 12-04-055JEE MainJEE Main 2025 (23 Jan, Shift 2)Easy
A galvanometer having a coil of resistance $30\ \Omega$ needs $20\ \text{mA}$ of current for full-scale deflection. If a maximum current of $3\ \text{A}$ is to be measured using this galvanometer, the resistance of the shunt to be added to the galvanometer should be $\dfrac{30}{X}\ \Omega$, where $X$ is
Answer: (B) $149$
The shunt carries the rest of the current, $I - I_g$, at the same voltage as the coil:
$$S = \frac{I_gG}{I - I_g} = \frac{0.02\times30}{3 - 0.02} = \frac{0.6}{2.98} = \frac{30}{149}\ \Omega$$
$X = 149$
Solution by Sreeraj P, M.Sc Physics